Solved Problems

Category: Integral

Solved Problems in Integrals

Problem 2-8: Integration

Find $\displaystyle\int \! \frac{e^x}{1+e^x}\, dx$ .

Solution

Let $u=1+e^x$ , therefore $du=e^x dx \to dx=\frac{1}{u-1}du$ . Hence

$\displaystyle\int \! \frac{e^x}{1+e^x}\, dx = \displaystyle\int \! \frac{u-1}{u} \frac{1}{u-1}\,du = \displaystyle\int \! \frac{1}{u} \,du=\ln (u)=\ln(1+e^x)$

April 10, 2010
Problem 2-7: Evaluatin Definite Integrals

Evaluate

a) $\displaystyle\int_{-1}^1 \! x^2 \, dx$
b) $\displaystyle\int_{-2}^1 \! \sin(x) \, dx$
c) $\displaystyle\int_{-1}^2 \! 2x^3+x-1 \, dx$
d) $\displaystyle\int_{-2}^2 \! |x| \, dx$

Solution
a) $\displaystyle\int_{-1}^1 \! x^2 \, dx=\frac{1}{3}x^3\Bigr|^1_{-1}=\frac{1}{3}(1)^3-\frac{1}{3}(-1)^3=\frac{2}{3}$
(more…)

April 10, 2010